# Show that the two straight lines x^2(tan^2theta+cos^2theta)-2xytantheta+y^2sin^2theta=0 makes angles with the x-axis such that the difference of their tangents is 2.?

Aug 5, 2018

The given equation of a pair of straight lines is
${x}^{2} \left({\tan}^{2} \theta + {\cos}^{2} \theta\right) - 2 x y \tan \theta + {y}^{2} {\sin}^{2} \theta = 0$
$\mathmr{and} , {y}^{2} - \frac{2 \tan \theta}{\sin} ^ 2 \theta x y + \frac{{\tan}^{2} \theta + {\cos}^{2} \theta}{\sin} ^ 2 \theta {x}^{2} = 0$
Let the equations of component straight lines are $y - {m}_{1} x = 0 \mathmr{and} y - {m}_{2} x = 0$, where ${m}_{1} \mathmr{and} {m}_{2}$ are the slopes of the two lines.

So we can write

$\left(y - {m}_{1} x\right) \left(y - {m}_{2} x\right) = {y}^{2} - \frac{2 \tan \theta}{\sin} ^ 2 \theta x y + \frac{{\tan}^{2} \theta + {\cos}^{2} \theta}{\sin} ^ 2 \theta {x}^{2}$
$\implies {y}^{2} - \left({m}_{1} + {m}_{2}\right) x y + {m}_{1} {m}_{2} {x}^{2} = {y}^{2} - \frac{2 \tan \theta}{\sin} ^ 2 \theta x y + \frac{{\tan}^{2} \theta + {\cos}^{2} \theta}{\sin} ^ 2 \theta {x}^{2}$

Comparing both sides we get

${m}_{1} + {m}_{2} = \frac{2 \tan \theta}{\sin} ^ 2 \theta = \frac{2 \sec \theta}{\sin} \theta$

And

${m}_{1} {m}_{2} = \frac{{\tan}^{2} \theta + {\cos}^{2} \theta}{\sin} ^ 2 \theta$

So

${\left({m}_{2} - {m}_{2}\right)}^{2} = {\left({m}_{1} + {m}_{2}\right)}^{2} - 4 {m}_{1} {m}_{2}$

$= {\left(\frac{2 \sec \theta}{\sin} \theta\right)}^{2} - \frac{4 \left({\tan}^{2} \theta + {\cos}^{2} \theta\right)}{\sin} ^ 2 \theta$

$= 4 \left[\frac{{\sec}^{2} \theta - {\tan}^{2} \theta - {\cos}^{2} \theta}{\sin} ^ 2 \theta\right]$

$= 4 \left[\frac{1 - {\cos}^{2} \theta}{\sin} ^ 2 \theta\right]$

$= 4 \left[\frac{{\sin}^{2} \theta}{\sin} ^ 2 \theta\right]$

$= 4$

Hence $\left\mid {m}_{1} - {m}_{2} \right\mid = 2$
So it proves that the difference of tangents of the angles made by the lines with the X-axis is 2